Question

Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)→3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. How much H2O2 was dissolved if the titration required 14.3 mL of the KMnO4 solution?

Answer 1

Answer:0.408 g amount of hydrogen peroxide was dissolved in 100 mL of water.

Explanation:

Molarity of the $KMnO_4$ solution = 1.68 M

$Molarity=\frac{\text{Number of moles(n)}}{\text{Volume in Liters}}$

$1.68=\frac{n}{0.0143 L}$

Number of moles of $KMnO_4$,n = 0.0240 moles

$2KMnO_4(aq)+H_2O_2(aq)+3H_2SO_4(aq)\rightarrow 3O_2(g)+2MnSO_4(aq)+K_2SO_4(aq)+4H_2O(l)$

According to reaction 2 moles of $KMnO_4$ reacts with 1 mole of $H_2O_2$, then  0.0240 moles of KMnO_4 will react with :$\frac{1}{2}\times 0.0240$ moles of $H_2O_2$ that is 0.0120 moles

Mass of $H_2O_2$=0.0120 moles × 34 g/mol=0.408 g

0.408 g amount of hydrogen peroxide was dissolved in 100 mL of water.